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Chapter 12: Problem 1698

Two batteries each of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series to a resistor \(R\). Maximum Possible power consumed by the resistor \(=\ldots .\) (A) \(3.2 \mathrm{~W}\) (B) \((16 / 9) \mathrm{W}\) (C) \((8 / 9) \mathrm{W}\) (D) \(2 \mathrm{~W}\)

Short Answer

Expert verified
The maximum possible power consumed by the resistor is \(2 \mathrm{~W}\), which corresponds to option (D).

Step by step solution

01

Calculate the total EMF and internal resistance of the circuit

The batteries are connected in series, so the total EMF of the circuit is the sum of the emf of both batteries, and the total internal resistance is the sum of their internal resistances. Total EMF (E_total) = E_1 + E_2 = 2V + 2V = 4V Total internal resistance (r_total) = r_1 + r_2 = 1Ω + 1Ω = 2Ω
02

Apply the maximum power theorem

To maximize the power consumed by the resistor, we need to find the value of R when the power is at its maximum. According to the maximum power transfer theorem, this occurs when the load resistance (R) is equal to the internal resistance (r_total) of the circuit. So, R = r_total = 2Ω
03

Calculate the total current in the circuit

Now, let's find the total current in the circuit (I_total) using Ohm's law: I_total = E_total / (R + r_total) = 4V / (2Ω + 2Ω) = 4V / 4Ω = 1A
04

Calculate the maximum power consumed by the resistor

Finally, let's calculate the maximum power (P_max) consumed by the resistor using the formula P = I^2 * R: P_max = I_total^2 * R = (1A)^2 * 2Ω = 1A * 2Ω = 2W So the maximum possible power consumed by the resistor is 2 W, which corresponds to option (D).

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